Verify using polar form that #i^9# is #i# Can someone please help me with this? I tried doing this and came up with z=r=#sqrt1#(cos90 + #i#sin90) And #z^9#=#sqrt1#^9(cos 810 + #i# sin 810)

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Answer 1 · 2017-03-26T13:18:48

See calculation below

#z=i=1(cos90+isin90)#

Applying De moivre's Theorem

#z^9=i^9=1^9(cos90+isin90)^9#

#i^9=1(cos810+isin810)#

#cos810=cos (90+2*360)=cos90=0#

#sin810=sin (90+2*360)=sin90=1#

So,

#i^9=0+i=i#

You were on the good path, you needed only to simplify

#cos810# and #sin810#

I hope this is helpful